Locus of centre of variable circle I am not able to figure out this question 
What is the locus of the centre of a circle which touches a given line and passes through a given point, not lying on the given line?
I think it's a parabola but I am not able to prove it mathematically
 A: Without loss of generality, we may assume that the line is the $x$-axis, and that the point is the point $(0,a)$, where $a$ is positive. 
A circle with centre $(p,q)$ that touches the $x$-axis must have radius $|q|$. So it has equation 
$$(x-p)^2+(y-q)^2=q^2.$$
The circle goes through $(0,a)$. It follows that
$$(0-p)^2+(a-q)^2=q^2.$$
Simplify. We get 
$$p^2+a^2-2aq=0.$$
We can rewrite this as 
$$q=\frac{1}{2a}p^2 +\frac{a}{2},$$
indeed the equation of a parabola. 
A: Assume the line is y-axis and the point is $(a ,0)$.the centre of circle is $(h,k)$ then the equation of circle:
$$(x-h)^2 +(y-k)^2 =r^2 \qquad (1)$$
where $r=$radius of  circle 
but $r=k$ 
putting this value in equation $(1)$
we get
$$(x-h)^2 +(y-k)^2 =k^2 \qquad (2)$$
but the circle is passing through the point $(a,0)$ therefore putting 
$x=a, y=0$ in equation $(2)$ we get 
$$(a-h)^2 +(0-k)^2 =k^2 $$
simplify:
$$a^2 +h^2 =2ah \qquad (3)$$
equation $(3)$ is the locus of centre of circle touching one axis and passing through a point not lying on the given line.
